Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int SIZE = 101;
int mat1[SIZE][SIZE] = {
    0,
};
int mat2[SIZE][SIZE] = {
    0,
};
vector<int> solveRow(int n, int m) {
  vector<int> ans;
  int i, j, minX, maxX;
  for (i = 1; i <= n; i++) {
    minX = 100000;
    for (j = 1; j <= m; j++) {
      minX = min(minX, mat1[i][j]);
    }
    for (j = 1; j <= m; j++) {
      mat1[i][j] -= minX;
      if (mat1[i][j] < 0) {
        ans.clear();
        return ans;
      }
    }
    while (minX--) {
      ans.push_back(i);
    }
  }
  for (j = 1; j <= m; j++) {
    minX = 100000;
    maxX = 0;
    for (i = 1; i <= n; i++) {
      minX = min(minX, mat1[i][j]);
      maxX = max(maxX, mat1[i][j]);
    }
    if (minX != maxX) {
      ans.clear();
      return ans;
    }
    for (i = 1; i <= n; i++) {
      mat1[i][j] -= minX;
      if (mat1[i][j] < 0) {
        ans.clear();
        return ans;
      }
    }
    while (minX--) {
      ans.push_back(j * -1);
    }
  }
  return ans;
}
vector<int> solveCol(int n, int m) {
  vector<int> ans;
  int i, j, minX, maxX;
  for (j = 1; j <= m; j++) {
    minX = 100000;
    for (i = 1; i <= n; i++) {
      minX = min(minX, mat2[i][j]);
    }
    for (i = 1; i <= n; i++) {
      mat2[i][j] -= minX;
      if (mat2[i][j] < 0) {
        ans.clear();
        return ans;
      }
    }
    while (minX--) {
      ans.push_back(j * -1);
    }
  }
  for (i = 1; i <= n; i++) {
    minX = 100000;
    maxX = 0;
    for (j = 1; j <= m; j++) {
      minX = min(minX, mat2[i][j]);
      maxX = max(maxX, mat2[i][j]);
    }
    if (minX != maxX) {
      ans.clear();
      return ans;
    }
    for (j = 1; j <= m; j++) {
      mat2[i][j] -= minX;
      if (mat2[i][j] < 0) {
        ans.clear();
        return ans;
      }
    }
    while (minX--) {
      ans.push_back(i);
    }
  }
  return ans;
}
int main(int argc, char* arg[]) {
  vector<int> ans, ansRow, ansCol;
  int n, m, i, j, x, globalMax;
  globalMax = 0;
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      cin >> x;
      mat1[i][j] = x;
      mat2[i][j] = x;
      globalMax = max(globalMax, x);
    }
  }
  if (globalMax == 0) {
    cout << "0\n";
  } else {
    ansRow = solveRow(n, m);
    ansCol = solveCol(n, m);
    if (ansRow.size() == 0 && ansCol.size() == 0) {
      cout << "-1\n";
      return 0;
    }
    if (ansCol.size() == 0 || ansRow.size() <= ansCol.size()) {
      ans = ansRow;
    } else {
      ans = ansCol;
    }
    cout << ans.size() << "\n";
    for (i = 0; i < ans.size(); i++) {
      if (ans[i] > 0) {
        cout << "row " << ans[i] << "\n";
      } else {
        cout << "col " << ans[i] * -1 << "\n";
      }
    }
  }
  return 0;
}